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Physics

A body starts with an initial velocity of 10 m s-1 and acceleration 5 m s-2. Find the distance covered by it in 5 s .

Motion in One Dimension

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Answer

According to equation of motion —

S = ut + 12\dfrac{1}{2} at2

Where,

distance covered = S

initial velocity u = 10 m s-1

acceleration a = 5 m s-2

time t = 5 s

Substituting the values in the formula, we get,

S=(10×5)+(12×5×52)S=(50)+(12×5×25)S=(50)+(2.5×25)S=50+62.5⇒S=112.5mS = (10 \times 5) + (\dfrac{1}{2} \times 5 \times 5^2) \\[0.5em] S = (50) + (\dfrac{1}{2} \times 5 \times 25) \\[0.5em] S = (50) + ( 2.5 \times 25) \\[0.5em] S = 50 + 62.5 \\[0.5em] \Rightarrow S = 112.5 m \\[0.5em]

Hence, distance covered = 112.5 m.

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